Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 105000;
int n, m, q;
struct edge {
  int v, next;
} E[maxn << 1];
int head[maxn], nedge;
inline void add(int u, int v) {
  E[++nedge].v = v;
  E[nedge].next = head[u];
  head[u] = nedge;
}
int val[maxn];
int p[maxn], hp[maxn], size[maxn], dpt[maxn], hs[maxn], order[maxn], ps[maxn];
inline void Begonia(int v0) {
  static int d[maxn], last, now[maxn], i, j, tot;
  size[v0] = dpt[d[last = 1] = v0] = 1;
  memcpy(now, head, sizeof head);
  while (last) {
    if (!(j = now[i = d[last]])) {
      if (--last) {
        size[d[last]] += size[i];
        if (size[i] > size[hs[d[last]]]) hs[d[last]] = i;
      }
      continue;
    }
    if (E[j].v != p[i])
      dpt[E[j].v] = dpt[p[E[j].v] = i] + (size[d[++last] = E[j].v] = 1);
    now[i] = E[j].next;
  }
  order[ps[v0] = tot = 1] = d[last = 1] = v0;
  memset(now, 0xFF, sizeof now);
  for (int i = 1; i <= n; ++i) hp[i] = i;
  while (last) {
    if (!(j = now[i = d[last]])) {
      --last;
      continue;
    }
    if (j == -1) {
      if (hs[i]) hp[order[ps[hs[i]] = ++tot] = d[++last] = hs[i]] = hp[i];
      now[i] = head[i];
      continue;
    }
    if (E[j].v != p[i] && E[j].v != hs[i])
      order[ps[E[j].v] = ++tot] = d[++last] = E[j].v;
    now[i] = E[j].next;
  }
}
inline void merge(vector<int> &lft, vector<int> &rgt, vector<int> &ans) {
  ans.clear();
  for (int l = 0, r = 0;
       (l < lft.size() || r < rgt.size()) && ans.size() < 10;) {
    if (l == lft.size()) {
      ans.push_back(rgt[r++]);
      continue;
    }
    if (r == rgt.size()) {
      ans.push_back(lft[l++]);
      continue;
    }
    if (lft[l] < rgt[r])
      ans.push_back(lft[l++]);
    else
      ans.push_back(rgt[r++]);
  }
}
struct zkw_segt {
  int base;
  vector<int> peo[280000];
  void preset() {
    for (base = 1; base < n + 2; base <<= 1)
      ;
  }
  void insert(int pos, int val) {
    vector<int> tmp1, tmp2;
    tmp1.push_back(val);
    for (int i = pos + base; i; i >>= 1) {
      merge(peo[i], tmp1, tmp2);
      peo[i] = tmp2;
    }
  }
  void query(int l, int r, vector<int> &ans) {
    vector<int> lft, rgt, tmp;
    for (int x = l + base - 1, y = r + base + 1; x < y - 1; x >>= 1, y >>= 1) {
      if (!(x & 1)) {
        merge(lft, peo[x ^ 1], tmp);
        lft = tmp;
      }
      if (y & 1) {
        merge(rgt, peo[y ^ 1], tmp);
        rgt = tmp;
      }
    }
    merge(lft, rgt, ans);
  }
} MEOW;
inline void query(int x, int y, vector<int> &ans) {
  int x0 = x, y0 = y;
  vector<int> lgt, rgt, tmp, tmp2;
  while (hp[x0] != hp[y0]) {
    if (dpt[hp[x0]] > dpt[hp[y0]]) {
      MEOW.query(ps[hp[x0]], ps[x0], tmp);
      merge(lgt, tmp, tmp2);
      lgt = tmp2;
      x0 = p[hp[x0]];
    } else {
      MEOW.query(ps[hp[y0]], ps[y0], tmp);
      merge(rgt, tmp, tmp2);
      rgt = tmp2;
      y0 = p[hp[y0]];
    }
  }
  if (ps[x0] > ps[y0]) swap(x0, y0);
  MEOW.query(ps[x0], ps[y0], tmp);
  merge(lgt, tmp, tmp2);
  lgt = tmp2;
  merge(lgt, rgt, ans);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  Begonia(1);
  MEOW.preset();
  for (int i = 1; i <= m; ++i) {
    int v;
    scanf("%d", &v);
    MEOW.insert(ps[v], i);
  }
  vector<int> ans;
  for (int i = 1; i <= q; ++i) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    query(u, v, ans);
    if (a > ans.size()) a = ans.size();
    printf("%d", a);
    for (int i = 0; i < a; ++i) printf(" %d", ans[i]);
    printf("\n");
  }
  return 0;
}